Answer:
The dart with the small mass will travel the farthest distance.
Explanation:
Acceleration is proportional to force times mass, and inertia is proportional to mass. Inertia is the reluctance of a moving body to stop, and a stationary body to start moving (inertia increses with mass). Assuming they both have the same aerodynamic design, and that they are both launched with the same force applied for the same time duration, the dart with less small mass will accelerate faster than the big mass dart. From this we can see that the small dart will have covered a longer distance before the effect of the force stops, when compared to the more massive dart.
Originally there must been
1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start
% = 2.25 / 150 = 1.5 % of 235 U left
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is 
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.